QUARTER FINAL STAGE 2024
Accra Academy: 45 points
Oyoko Methodist SHS: 36 points
St. Joseph Seminary SHS: 28 points
ROUND 1
SECOND SET OF QUESTIONS
PREAMBLE
An object weighs $360\text{ N}$ on the surface of the earth.
Find the weight of the object at the given height above the earth.
FORMULA
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
FORMULA 2
Given radius of earth: $\dfrac{a}{b}$
The formula becomes $W = W_{0}\left(\dfrac{b}{a + b}\right)^{2}$
1. One fifth radius of earth
ANSWER: $250\text{ N}$
SOLUTION 1:
Given values:
$W_{0} = 360\text{ N}$
$h = \dfrac{R}{5}$
Substitution:
$W = 360 \times \left(\dfrac{R}{R + \dfrac{R}{5}}\right)^{2}$
$W = 360 \times \left(\dfrac{R}{\dfrac{6R}{5}}\right)^{2}$
$W = 360 \times \left(\dfrac{5}{6}\right)^{2}$
$W = 360 \times \dfrac{25}{36}$
$W = 10 \times 25$
$W = 250\text{ N}$
SOLUTION 2:
Radius: $\dfrac{a}{b} = \dfrac{1}{5}$
$W = W_{0}\left(\dfrac{b}{a + b}\right)^{2}$
$W = W_{0}\left(\dfrac{5}{1 + 5}\right)^{2}$
$W = 360 \times \left(\dfrac{5}{6}\right)^{2}$
$W = 360 \times \dfrac{25}{36}$
$W = 10 \times 25$
$W = 250\text{ N}$
2. Half the radius of earth
ANSWER: $160\text{ N}$
SOLUTION 1:
Given values:
$W_{0} = 360\text{ N}$
$h = \dfrac{R}{2}$
Substitution:
$W = 360 \times \left(\dfrac{R}{R + \dfrac{R}{2}}\right)^{2}$
$W = 360 \times \left(\dfrac{R}{\dfrac{3R}{2}}\right)^{2}$
$W = 360 \times \left(\dfrac{2}{3}\right)^{2}$
$W = 360 \times \dfrac{4}{9}$
$W = 40 \times 4$
$W = 160\text{ N}$
SOLUTION 2:
Radius: $\dfrac{a}{b} = \dfrac{1}{2}$
$W = W_{0}\left(\dfrac{b}{a + b}\right)^{2}$
$W = W_{0}\left(\dfrac{2}{1 + 2}\right)^{2}$
$W = 360 \times \left(\dfrac{2}{3}\right)^{2}$
$W = 360 \times \dfrac{4}{9}$
$W = 40 \times 4$
$W = 160\text{ N}$
3. One earth radius
ANSWER: $90\text{ N}$
SOLUTION:
Given values:
$W_{0} = 360\text{ N}$
$h = R$
Substitution:
$W = 360 \times \left(\dfrac{R}{R + R}\right)^{2}$
$W = 360 \times \left(\dfrac{R}{2R}\right)^{2}$
$W = 360 \times \left(\dfrac{1}{2}\right)^{2}$
$W = 360 \times \dfrac{1}{4}$
$W = 90\text{ N}$
SOLUTION 2:
Radius: $\dfrac{a}{b} = \dfrac{1}{1}$
$W = W_{0}\left(\dfrac{b}{a + b}\right)^{2}$
$W = W_{0}\left(\dfrac{1}{1 + 1}\right)^{2}$
$W = 360 \times \left(\dfrac{1}{2}\right)^{2}$
$W = 360 \times \dfrac{1}{4}$
$W = 90\text{ N}$
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PRACTICE QUESTIONS
1. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of two earth radii above the surface.
ANSWER: 64.0 N or $6.40\times10^{1}$ N
SOLUTION 1:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + 2R}\right)^{2}$
$W = 576 \times \dfrac{1}{9}$
$W = 64.0\text{ N}$
SOLUTION 2:
Radius: $\dfrac{a}{b} = \dfrac{2}{1}$
$W = W_{0}\left(\dfrac{b}{a + b}\right)^{2}$
$W = W_{0}\left(\dfrac{1}{1 + 2}\right)^{2}$
$W = 576 \times \dfrac{1}{9}$
$W = 64.0\text{ N}$
2. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of five earth radii above the surface.
ANSWER: 16.0 N or $1.60\times10^{1}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + 5R}\right)^{2}$
$W = 576 \times \dfrac{1}{36}$
$W = 16.0\text{ N}$
3. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of three earth radii above the surface.
ANSWER: 36.0 N or $3.60\times10^{1}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + 3R}\right)^{2}$
$W = 576 \times \dfrac{1}{16}$
$W = 36.0\text{ N}$
4. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of eleven earth radii above the surface.
ANSWER: 4.00 N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + 11R}\right)^{2}$
$W = 576 \times \left(\dfrac{1}{12}\right)^{2}$
$W = 576 \times \dfrac{1}{144}$
$W = 4.00\text{ N}$
5. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of seven earth radii above the surface.
ANSWER: 9.00 N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + 7R}\right)^{2}$
$W = 576 \times \dfrac{1}{64}$
$W = 9.00\text{ N}$
6. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of nine earth radii above the surface.
ANSWER: 5.76 N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + 9R}\right)^{2}$
$W = 576 \times \dfrac{1}{100}$
$W = \dfrac{144}{25}$
$W = 5.76\text{ N}$
7. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of a seventh of the earth's radius above the surface.
ANSWER: 441 N or $4.41\times10^{2}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + \dfrac{R}{7}}\right)^{2}$
$W = 576 \times \left(\dfrac{7}{8}\right)^{2}$
$W = 576 \times \dfrac{49}{64}$
$W = 9 \times 49$
$W = 441\text{ N}$
8. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of a third of the earth's radius above the surface.
ANSWER: 324 N or $3.24\times10^{2}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + \dfrac{R}{3}}\right)^{2}$
$W = 576 \times \left(\dfrac{3}{4}\right)^{2}$
$W = 576 \times \dfrac{9}{16}$
$W = 9 \times 36$
$W = 324\text{ N}$
9. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of one earth radius above the surface.
ANSWER: 144 N or $1.44\times10^{2}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + R}\right)^{2}$
$W = 576 \times \left(\dfrac{1}{2}\right)^{2}$
$W = 576 \times \dfrac{1}{4}$
$W = 144\text{ N}$
10. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of five-thirds of the earth's radius above the surface.
ANSWER: 81.0 N or $8.10\times10^{1}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + \dfrac{5R}{3}}\right)^{2}$
$W = 576 \times \left(\dfrac{3}{8}\right)^{2}$
$W = 576 \times \dfrac{9}{64}$
$W = 9 \times 9$
$W = 81.0\text{ N}$
11. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of half the earth's radius above the surface.
ANSWER: 256 N or $2.56\times10^{2}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + \dfrac{R}{2}}\right)^{2}$
$W = 576 \times \left(\dfrac{2}{3}\right)^{2}$
$W = 576 \times \dfrac{4}{9}$
$W = 64 \times 4$
$W = 256\text{ N}$
12. QUESTION: An object weighs 576 N on the surface of the earth. Find its weight at a height of three-fifths of the earth's radius above the surface.
ANSWER: 225 N or $2.25\times10^{2}$ N
SOLUTION:
$W = W_{0}\left(\dfrac{R}{R+h}\right)^{2}$
$W = 576 \times \left(\dfrac{R}{R + \dfrac{3R}{5}}\right)^{2}$
$W = 576 \times \left(\dfrac{5}{8}\right)^{2}$
$W = 576 \times \dfrac{25}{64}$
$W = 25 \times 9$
$W = 225\text{ N}$